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following Tito Piezas III comment I'll provide some concrete examples. First, let's take the second convergent of the cfrac $$G(q,k)\approx\cfrac{1}{1-q+\cfrac{1}{1-{q^3}^k}}$$ and expand it into a power series as $k\to\infty$ $$G(q,k)\approx \frac{1}{2}+\frac{1}{2^2}q+\frac{1}{2^3}q^2+\dots$$ Which converges to $\frac{2}{3}$ for $q=\frac{1}{2}$ as shown ...

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Take $k=1$ and $q=1/2.$ Then the first few convergents are (starting with the trivial zeroth which here is $0$) $$C(0)=0,\\ C(1)=2,\\ C(2)=14/23=0.60869..,\\ C(3)=946/969=0.97626..,\\ C(5)=177486/217271=0.81688.$$ Since the terms $b(k)=1-(1/2)^{2k-1}$ go so rapidly to $1$ it would seem odd (to me) if the convergents were not alternately below/above the ...

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The difference is that Alpha is giving you a negative continued fraction-note that it is $- [1; 1, 8, 1, 3, 2, 2, 9, \dots]$ Your answer key is giving you a positive continued fraction that is added to $-2$. If you add $3106$ to the numerator to make the value positive, Alpha gives $[0; 9, 1, 3, \overline{2, 2, 9, 1, 1, 2, 1, 4, 13, 9, 1, 9, 1, 118, 1, 9, ... 5 The answer is yes. Given the nome$q = \exp(i\pi\tau)$, elliptic lambda function$\lambda(\tau)$, Dedekind eta function$\eta(\tau)$, Jacobi theta functions$\vartheta_n(0,q), and Ramanujan's octic cfrac, the following relations are known, \begin{aligned} u(\tau) & = \big(\lambda(\tau)\big)^{1/8} = \frac{\sqrt{2}\, \eta(\tfrac{\tau}{2})\, ... 2 So with formal manipulation:\psi(1/q)= \frac{-q^{-1}}{1-q^{-1}+ \frac{q^{-1}(1-q^{-1})^2}{1-q^{-3}+\ldots}}$$Times top and bottom by -q to give:$$\psi(1/q)= \frac{1}{1-q+ \frac{-qq^{-3}(1-q)^2}{1-q^{-3}+\frac{q^{-1}(1-q^{-2})^2}{1-q^{-5}+\ldots}}}$$Then mutiply top and bottom of the next fraction by -q^3 to give:$$\psi(1/q)= \frac{1}{1-q+ ... 2 To clarify, what you found is a q-continued fraction for the Jacobi theta function\vartheta_2(0,q). Using ccorn's formulation, $$\left(\frac{\vartheta_2(0,q)}{2\,q^{1/4}}\right)^2 =\Big(\sum_{n=0}^\infty q^{n(n+1)}\Big)^2 ... 1 Regarding R. Israel's remark, three of your continued fractions, while not exactly the same, are variants of a common form discussed here for |q|<1,$$\frac{1}{1-q} =\cfrac{1}{1+q-\cfrac{\color{brown}{2q(1+q^2)}}{1+q^3+\cfrac{q^2(1-q)(1-q^3)}{1+q^5-\cfrac{q^3(1+q^2)(1+q^4)}{1+q^7+\cfrac{q^4(1-q^3)(1-q^5)}{1+q^9-\ddots}}}}}\tag0$$First, the one for ... 1 This supplements R. Israel's answer. Given the continued fraction discussed in this post for |q|<1,$$\frac{1}{1-q} =\cfrac{1}{1+q-\cfrac{\color{brown}{2q(1+q^2)}}{1+q^3+\cfrac{q^2(1-q)(1-q^3)}{1+q^5-\cfrac{q^3(1+q^2)(1+q^4)}{1+q^7+\cfrac{q^4(1-q^3)(1-q^5)}{1+q^9-\ddots}}}}}$$and using a little algebraic manipulation to transform the brown part to ... 2 (A partial answer.) This is a special case of a conjectured equality discussed in this MO post. Let |q|<1, then,$$\begin{aligned}U(q) &= \prod_{n=0}^\infty \frac{\big(1-a^2q^3(q^4)^n\big)\big(1-b^2q^3(q^4)^n\big)}{\big(1-a^2q(q^4)^n\big)\big(1-b^2q(q^4)^n\big)}\\ &= \dfrac{1} {1+ab-\dfrac{(a+bq)(b+aq)} {1+(ab)^3+\dfrac{(a-bq^2)(b-aq^2)q} ... 3 The ordinary generating function for your recurrence is $$g(x) = \dfrac{1+x^2}{1-x-x^3}$$ Thus $$\sum_{n=0}^\infty (-1)^n a_n q^n = g(-q) = \frac{1+q^2}{1+q+q^3}\tag1$$ If that is\phi(q), then indeed $$\phi(1/q) = \dfrac{1/q^2+1}{1/q^3 + 1/q + 1} = \dfrac{q (1 + q^2)}{1+q^2 + q^3} = q \phi(q)$$ Now let's try to get your continued fraction. ... 2 Let F(x)=x^0+\cfrac1{x^1+\cfrac1\cdots}~. Then F(2) is OEIS A214070, for which no closed form is currently known. 1 You want the value which is greater than 1 - clearly by estimating. From x^8+\frac 1{x^8}=2207 you know that if x satisfies that equation so will \frac 1x. Put \frac 1x in the original equation and modify the eighth root accordingly and you will see where the alternative answer comes from and that this is clearly less than 1. 3 The 8th root is not pertinent. You are asking how to assign a value to the continued fraction inside the 8th root. As the equation for x^8 shows, there are two solutions of f = 2207 - \frac{1}{f}. My first thought was that the even and odd "levels" of the continued fraction converge to the two different solutions (making it hard to distinguish the ... 0 Note: OP's expression is usually regarded as continued fraction. We will show it has the single solution 2. Before we analyse OPs expression, let's have a look at a continued fraction representation of \sqrt{2} \begin{align*} \sqrt{2}=[1;2,2,2,\ldots]=1+\frac{1}{2+\frac{1}{2+\frac{1}{2+\ddots}}} \end{align*} The convenient notation ... 2 Note that as we add more terms to the continued fraction, it oscillates between 1 and slightly higher than 2. \begin{align} n&=1& 3&=3& 3-2&=1\\\\ n&=2& 3-\cfrac23&=\frac73& 3-\cfrac{2}{3-2}&=1\\\\ n&=3& 3-\cfrac{2}{3-\cfrac23}&=\frac{15}7& 3-\cfrac2{3-\cfrac2{3-2}}&=1\\\\ n&=4& ... 5 Let us define two series. The first is \begin{align} a_1 &= 3 \\ a_2 &= 3 - \frac{2}{3} \\ a_3 &= 3 - \frac{2}{3 - \frac{2}{3}} \\ a_4 &= 3- \frac{2}{3 - \frac{2}{3 - \frac{2}{3}}} \\ &\vdots \\ a_{n+1} &= 3 - \frac{2}{a_n} \quad (*) \end{align} and \begin{align} b_1 &= 3 - 2 \\ b_2 &= 3 - \frac{2}{3-2} \\ b_3 &= 3 - ... 6 This continued fraction is the limit of the sequencea_n=3-2/a_{n-1}$. Computing the first few terms shows that$2$is the correct limit; if our initial term$a_1=3$were different then the limit could be$1$. -1 Your two solutions are correct. You can verify them as follows. $$\color{blue}{1} = 3 - 2=3-\frac{2}{\color{\red}{1}}\tag{1}$$ Now replace the red 1 with the blue 1, which equals to the right hand side in (1). $$\color{blue}{2} = 3 - 1=3-\frac{\color{cyan}{2}}{\color{\red}{2}}\tag{2}$$ Now replace the red 2 with the blue 2, which equals to the right hand ... 2 Look at it as two different series and you will understand why both 1 and 2 are possible solutions of this: series 1:$\{3-2, 3-\frac{2}{3-2}, 3-\frac{2}{3-\frac{2}{3-2}}, ...\}$series 2:$\{3-\frac{2}{3}, 3-\frac{2}{3-\frac{2}{3}}, 3-\frac{2}{3-\frac{2}{3-\frac{2}{3}}}, ...\}$. series 1 converges to 1 whereas series 2 converges to 2. 6 If you continue adding numbers to the expression one at a time, then you have the sequence $$3,\; 3-2,\; 3-\frac{2}{3},\; 3-\frac{2}{3-2},\; 3-\frac{2}{3-\frac{2}{3}},\; 3-\frac{2}{3-\frac{2}{3-2}},\;3-\frac{2}{3-\frac{2}{3-\frac{2}{3}}}\ldots,$$ or $$3,\; 1,\; \frac{7}{3},\; 1,\; \frac{15}{7},\; 1,\;\frac{31}{15},\;\ldots,$$ which consists of two ... 1 Also, if we take the house numbers $$x_0 = 0,$$ $$x_1 = 1,$$ $$x_2 = 6,$$ $$x_3 = 35,$$ $$x_4 = 204,$$ $$x_5 = 1189,$$ we have a simple recurrence $$\color{magenta}{ x_{n+2} = 6 x_{n+1} - x_n }$$ which follows by Cayley-Hamilton from the generator of the oriented automorphism group of$u^2 - 8 v^2.$If$y_n$is the number of houses on the ... 2 If it is house number$x$in a street of$y$houses, we have $$\frac{x(x-1)}{2}+x+\frac{x(x-1)}{2}=\frac{y(y+1)}{2}$$ which simplifes to $$(2y+1)^2-8x^2=1\ .$$ This can be solved by computing the continued fraction $$\sqrt8=2+\frac{1}{1+{}}\frac{1}{4+}\frac{1}{1+{}}\frac{1}{4+\cdots}\ .$$ The table of convergents is ... 4 I will stick to verifying the identity numerically. A backwards recursion formula for the$n$'th partial quotient of the continued fraction is $$s_{k-1} = 1 - e^{-(2k-1)\pi} + \frac{e^{-k\pi}(1+e^{-k\pi})^2}{s_{k}}$$ for$k=n,n-1,\ldots,3,2,1$and$s_n = 1$. Having calculated$s_0$the$n$'th partial quotient is then given as$1 + \frac{2e^{-\pi/2}}{s_0}\$. ...

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